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julsineya [31]
3 years ago
8

what is equal and opposite, according to Newton's third law? action change in acceleration reaction change in velocity

Physics
1 answer:
Finger [1]3 years ago
4 0

Answer: action and reaction

Explanation: Newton's 3rd states that action and reaction are equal and opposite in direction

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Starting at (0,0) an object travels 36 meters north and then it covers 20 meters east. What is
Svetradugi [14.3K]

Answer:

Explanation:

Using the pythagoras theorem, the displacement is expressed as;

d² = x²+y²

y = 36m (north)

x = 20m east

Substitute;

d² = 36²+20²

d² = 1296+400

d² = 1696

d = √1696

d = 41.18m

For the direction;

theta = tan^-1(y/x)

theta = tan^-1(36/20)

theta = tan^-1(1.8)

theta = 60.95°

Hence the magnitude is 41.18m and the direction is 60.95°

8 0
3 years ago
If you can answer all of this then your a legend (I'm giving you all my points)
navik [9.2K]

Answer:

1. B

2. B

3. D

4. A

Hope this was correct! A lot of the answers are already in the article itself and the wording is just different. I suggest now that for information retainment, you read the article again with the correct points in mind and see if you can spot the points where the answers are stated!

3 0
2 years ago
Which statement describes the energy transformation that occurs when a person eats a sandwich to gain energy for a long hike? O
Nataly [62]

Answer:chemical energy is transformed into kinetic energy

Explanation:

7 0
3 years ago
Every winter I fly to Michigan. It's a total distance of 3900km. It takes 5 hours. What is my average speed?
olga_2 [115]
\frac{3900}{5} = 780\sf ~kilometer~per~hour

Your average speed is 780 kilometer per hour.
4 0
3 years ago
Read 2 more answers
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
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